How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s

The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4

scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

v = v₀ + a t

y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

Positive slope the body is accelerating and the speed is increasing.

Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s

v = v₀ + a t

$a = \frac{v-v_o}{\Delta t}$

$a = \frac{2-0}{8-0}$

a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

$a = \frac{6-2}{ 14-10}$

a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

$d_{total } = d_1 +d_2 + d_3 +d_4$

We look for every displacement.

d₁ = v₀ + ½ a₁ Δt²

d₁ = 0 + ½ 0.25 8²

d₁ = 8 m

In the second section the velocity is constant

d₂ = v₂ Δt₂

d₂ = 2 (10-8)

d₂ = 4 m

The third section.

d₃ = v₀ + ½ a t²

d₃ = 2 + ½ 1 (14-10) ²

d₃ = 10 m

The distance of the fourth section.

we look for acceleration

a₄ = $\frac{v-v_o}{\Delta t}$

a₄ = $\frac{0-6}{18-14}$

a₄ = -1.5 m / s²

d₄ = 6 + ½ (-1.5) (1814) ²

d₄ = -6 m

The total displacement is;

$d_{total}$ = 8 + 4 + 10 -6

$d_{total}$ = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0

2 years ago

Plz tll answer of this . pleaseeeeeeeeeeee. Q 6,7,8<br>

777dan777 [17]

Can someone pls help us with this question I need the answer too

8 0

3 years ago

This diagram shows the process that powers stars. This process is called?

viktelen [127]

The process that powers stars is C)fusion

4 0

3 years ago

Read 2 more answers

What is the difference between real and apparent weightlessness?

MrRissso [65]

Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

4 0

3 years ago

A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a

Digiron [165]

The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>

We have the expression for torque acting on a current loop in a uniform magnetic field as,

$\tau=MBsin\theta$

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
Thus, the potential energy will be,

$\tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J$

Thus, we can conclude that, the potential energy will be 1.46*10^-4J.

Learn more about the torque here:

brainly.com/question/27949876

#SPJ4

7 0

2 years ago